Answer
See below
Work Step by Step
Given:
$$(x^5+y^m)dx-x^ny^3dy=0$$
a) For our equation:
$M_y'=my^{m-1}\\
N'_x=-nx^{n-1}y^3$
The only solution that satisfies this condition is: $m=n=0$
b) Rewrite the given differential equation in the equivalent form:
$\frac{dy}{dx}=\frac{x^5+y^m}{x^ny^3}$
then $m=0,n \in R$
c) To get a homogeneous equation from the given equation, we rewrite:
$\frac{dy}{dx}=\frac{x^5}{x^ny^3}+\frac{x^m}{x^ny^3}$
Thus, the solution to this is: $m=5\\n=2$
d) This is a Bernoulli equation where
$\frac{dy}{dx}-x^{-n}y^{m-3}=x^{5-n}y^{-3}$
Divide both sides by $y^{-3}$ we have:
$\frac{1}{y^3}\frac{dy}{dx}-x^{-n}y^{m}=x^{5-n}$
There is no value for $m$ or $n$ for which our equation is linear.
e) From (d), we get: $\frac{dy}{dx}-x^{-n}y^{m-3}=x^{5-n}y^{-3}$
The value that satisfies the condition for the Benourlli equation is: $m=4$
and $n$ can be arbitrary number.
